Is sqrt(2)cos(x-pi/4)=cos(x)+sin(x) ?

The answer to whether sqrt(2)cos(x-pi/4)=cos(x)+sin(x) is True

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Popular >

prove sqrt(2)cos(x-pi/4)=cos(x)+sin(x)

Solution

prove

2 cos (x - \frac{π}{4}) = cos (x) + sin (x)

Solution

True

Solution steps

2 cos (x - \frac{π}{4}) = cos (x) + sin (x)

Manipulating left side

2 cos (x - \frac{π}{4})

Rewrite using trig identities

cos (x - \frac{π}{4})

Use the Angle Difference identity:

cos (s - t) = cos (s) cos (t) + sin (s) sin (t)

= cos (x) cos (\frac{π}{4}) + sin (x) sin (\frac{π}{4})

Simplify

cos (x) cos (\frac{π}{4}) + sin (x) sin (\frac{π}{4}) : \frac{2 cos ( x ) + 2 sin ( x )}{2}

cos (x) cos (\frac{π}{4}) + sin (x) sin (\frac{π}{4})

cos (x) cos (\frac{π}{4}) = \frac{2 cos ( x )}{2}

cos (x) cos (\frac{π}{4})

Simplify

cos (\frac{π}{4}) : \frac{2}{2}

cos (\frac{π}{4})

Use the following trivial identity:

cos (\frac{π}{4}) = \frac{2}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

= \frac{2}{2} cos (x)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 cos ( x )}{2}

sin (x) sin (\frac{π}{4}) = \frac{2 sin ( x )}{2}

sin (x) sin (\frac{π}{4})

Simplify

sin (\frac{π}{4}) : \frac{2}{2}

sin (\frac{π}{4})

Use the following trivial identity:

sin (\frac{π}{4}) = \frac{2}{2}

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

= \frac{2}{2} sin (x)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 sin ( x )}{2}

= \frac{2 cos ( x )}{2} + \frac{2 sin ( x )}{2}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{2 cos ( x ) + 2 sin ( x )}{2}

= \frac{2 cos ( x ) + 2 sin ( x )}{2}

= 2 \frac{2 cos ( x ) + 2 sin ( x )}{2}

Simplify

2 \frac{2 cos ( x ) + 2 sin ( x )}{2} : cos (x) + sin (x)

2 \frac{2 cos ( x ) + 2 sin ( x )}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( 2 cos ( x ) + 2 sin ( x ) ) 2}{2}

Cancel

\frac{( 2 cos ( x ) + 2 sin ( x ) ) 2}{2} : \frac{2 cos ( x ) + 2 sin ( x )}{2}

\frac{( 2 cos ( x ) + 2 sin ( x ) ) 2}{2}

Apply radical rule:

n a = a^{\frac{1}{n}}

2 = 2^{\frac{1}{2}}

= \frac{2 ^{\frac{1}{2}} ( 2 cos ( x ) + 2 sin ( x ) )}{2}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{2 ^{\frac{1}{2}}}{2 ^{1}} = \frac{1}{2 ^{1 - \frac{1}{2}}}

= \frac{2 cos ( x ) + 2 sin ( x )}{2 ^{1 - \frac{1}{2}}}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= \frac{2 cos ( x ) + 2 sin ( x )}{2 ^{\frac{1}{2}}}

Apply radical rule:

a^{\frac{1}{n}} = n a

2^{\frac{1}{2}} = 2

= \frac{2 cos ( x ) + 2 sin ( x )}{2}

= \frac{2 cos ( x ) + 2 sin ( x )}{2}

Factor out common term

2

= \frac{2 ( cos ( x ) + sin ( x ) )}{2}

Cancel the common factor:

2

= cos (x) + sin (x)

= cos (x) + sin (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is sqrt(2)cos(x-pi/4)=cos(x)+sin(x) ?
The answer to whether sqrt(2)cos(x-pi/4)=cos(x)+sin(x) is True